Chung minh rang voi moi goc luong giac α lam cho bieu thuc xac dinh thi
a) \(\dfrac{1-sin2\alpha}{1+sin2\alpha}\)=cot\(^2\)(\(\dfrac{\pi}{4}\)+α) b) \(\dfrac{sin\alpha+sin\beta cos\left(\alpha+\beta\right)}{cos\alpha-sin\beta sin\left(\alpha+\beta\right)}\)=tan\(\left(\alpha+\beta\right)\).
a, \(\dfrac{1-sin2a}{1+sin2a}\)
\(=\dfrac{sin^2a+cos^2a-2sina.cosa}{sin^2a+cos^2a+2sina.cosa}\)
\(=\dfrac{\left(sina-cosa\right)^2}{\left(sina+cosa\right)^2}\)
\(=\dfrac{2sin^2\left(a-\dfrac{\pi}{4}\right)}{2sin^2\left(a+\dfrac{\pi}{4}\right)}\)
\(=\dfrac{sin^2\left(\dfrac{\pi}{4}-a\right)}{sin^2\left(a+\dfrac{\pi}{4}\right)}\)
\(=\dfrac{cos^2\left(\dfrac{\pi}{4}+a\right)}{sin^2\left(\dfrac{\pi}{4}+a\right)}=cot\left(\dfrac{\pi}{4}+a\right)\)
b, \(\dfrac{sina+sinb.cos\left(a+b\right)}{cosa-sinb.sin\left(a+b\right)}\)
\(=\dfrac{sina+sinb.cosa.cosb-sinb.sina.sinb}{cosa-sinb.sina.cosb-sinb.cosa.sinb}\)
\(=\dfrac{sina.\left(1-sin^2b\right)+sinb.cosa.cosb}{cosa.\left(1-sin^2b\right)-sinb.sina.cosb}\)
\(=\dfrac{sina.cos^2b+sinb.cosa.cosb}{cosa.cos^2b-sinb.sina.cosb}\)
\(=\dfrac{\left(sina.cosb+sinb.cosa\right).cosb}{\left(cosa.cosb-sinb.sina\right).cosb}\)
\(=\dfrac{sin\left(a+b\right)}{cos\left(a+b\right)}=tan\left(a+b\right)\)
1) Goi alpha la so do cua 1 cung luong giac co diem dau A, diem cuoi B. Khi do so do cua cac cung luong giac bat ki co diem dau A, diem cuoi B bang
A. pi - alpha + k2pi, k thuoc Z
B. alpha + kpi, k thuoc Z
C. alpha + k2pi, k thuoc Z
D. -alpha + k2pi, k thuoc Z
chung minh voi voi goc nhon \(\alpha\)tuy y ta co
\(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}\)
\(\cos\alpha=\frac{\cos\alpha}{\sin\alpha}\)
\(\tan\alpha.\cos\alpha=1\)
\(\sin^2\alpha+\cos^2\alpha=1\)
goi y: su dung dinh ly pitago
1) \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}\)
bien doi ve phai ta co
\(\frac{\sin\alpha}{\cos\alpha}=\frac{doi}{huyen}:\frac{ke}{huyen}=\frac{doi}{huyen}.\frac{huyen}{ke}=\frac{doi}{ke}=\tan\alpha\)
2) \(\cot\alpha=\frac{\cos\alpha}{\sin\alpha}\)
bien doi ve phai ta co
\(\frac{\cos\alpha}{\sin\alpha}=\frac{ke}{huyen}:\frac{doi}{huyen}=\frac{ke}{huyen}.\frac{huyen}{doi}=\frac{ke}{doi}=\cot\alpha\)
3) \(\tan\alpha.\cos\alpha=1\)
\(\frac{\cos\alpha}{\sin\alpha}.\frac{\sin\alpha}{\cos\alpha}=1\)
4) \(\sin^2\alpha+\cos^2\alpha=1\)
\(\frac{doi^2}{huyen^2}+\frac{ke^2}{huyen^2}=\frac{huyen^2}{huyen^2}=1\)( su dung dinh ly pitago )
rút gọn biểu thức
K= \(\frac{sin\left(\alpha+\beta\right)+sin\alpha+sin\beta}{cos\left(\alpha+\beta\right)+cos\alpha+cos\beta+1}\)
\(K=\frac{2sin\left(\frac{a+b}{2}\right).cos\left(\frac{a+b}{2}\right)+2sin\left(\frac{a+b}{2}\right).cos\left(\frac{a-b}{2}\right)}{2cos^2\left(\frac{a+b}{2}\right)-1+2cos\left(\frac{a+b}{2}\right).cos\left(\frac{a-b}{2}\right)+1}\)
\(K=\frac{sin\left(\frac{a+b}{2}\right)\left[cos\left(\frac{a+b}{2}\right)+cos\left(\frac{a-b}{2}\right)\right]}{cos\left(\frac{a+b}{2}\right)\left[cos\left(\frac{a+b}{2}\right)+cos\left(\frac{a-b}{2}\right)\right]}\)
\(K=\frac{sin\left(\frac{a+b}{2}\right)}{cos\left(\frac{a+b}{2}\right)}=tan\left(\frac{a+b}{2}\right)\)
H= \(\sin^4\alpha\left(1+2\cos^2\alpha\right)+\cos^4\alpha\left(1+2\sin^2\alpha\right)\)
K=\(\tan^2\alpha\left(2\cos^2\alpha+\sin^2\alpha-1\right)\)
Rút gọn biểu thức
Tính:
a, \(\sin32^0-\cos68^0\).
b, \(1-\cos^2\)α..
c, \((1-\sin\)α)(\(1-\sin\)α).
d, \(1+\cos^2\)α + \(\sin^2\)α.
e, Sinα - sinα . cos2α.
f, Sin2α + 2sinα . cosα + cos2α.
g, Tan2α - sin2α . tan2α.
h, Cos2α + tan2α . cos2α.
i, Tan2α ( 2cos2α + sin2α - 1).
k, Sin250 + sin2250 + sin2650 + sin2850 - 2.
RÚT GỌN:
a, \(A=\dfrac{\left(\cos\alpha-\sin\alpha\right)^2-\left(\cos\alpha-\sin^2\alpha\right)}{\cos\alpha.\sin\alpha}\)
\(b,B=\sin^6\alpha+\cos^6\alpha+3\sin^6\alpha.\cos^2\alpha\)
1. Cho tam giác $ABC$. Chứng minh rằng $\sin ^{2} A+\sin ^{2} B-\sin ^{2} C=2\sin A.\sin B.\cos C$.
2. Chứng minh rằng:
a. $\sin \alpha .\sin \left(\dfrac{\pi }{3} -\alpha \right).\sin \left(\dfrac{\pi }{3} +\alpha \right)=\dfrac{1}{4} \sin 3\alpha $
b. $\sin 5\alpha -2\sin \alpha \left({\rm cos} {\rm 4}\alpha +\cos 2\alpha \right)=\sin \alpha $
1/ Cho \(cot\alpha=\sqrt{5}\) . Tính \(C=sin^2\alpha-sin\alpha cos\alpha+cos^2\alpha\)
2/ Cho \(tan\alpha=3\) . Tính \(B=\dfrac{sin\alpha-cos\alpha}{sin^3\alpha+3cos^3\alpha+2sin\alpha}\)
1) \(cot\alpha=\sqrt[]{5}\Rightarrow tan\alpha=\dfrac{1}{\sqrt[]{5}}\)
\(C=sin^2\alpha-sin\alpha.cos\alpha+cos^2\alpha\)
\(\Leftrightarrow C=\dfrac{1}{cos^2\alpha}\left(tan^2\alpha-tan\alpha+1\right)\)
\(\Leftrightarrow C=\left(1+tan^2\alpha\right)\left(tan^2\alpha-tan\alpha+1\right)\)
\(\Leftrightarrow C=\left(1+\dfrac{1}{5}\right)\left(\dfrac{1}{5}-\dfrac{1}{\sqrt[]{5}}+1\right)\)
\(\Leftrightarrow C=\dfrac{6}{5}\left(\dfrac{6}{5}-\dfrac{\sqrt[]{5}}{5}\right)=\dfrac{6}{25}\left(6-\sqrt[]{5}\right)\)
1: \(cota=\sqrt{5}\)
=>\(cosa=\sqrt{5}\cdot sina\)
\(1+cot^2a=\dfrac{1}{sin^2a}\)
=>\(\dfrac{1}{sin^2a}=1+5=6\)
=>\(sin^2a=\dfrac{1}{6}\)
\(C=sin^2a-sina\cdot\sqrt{5}\cdot sina+\left(\sqrt{5}\cdot sina\right)^2\)
\(=sin^2a\left(1-\sqrt{5}+5\right)=\dfrac{1}{6}\cdot\left(6-\sqrt{5}\right)\)
2: tan a=3
=>sin a=3*cosa
\(1+tan^2a=\dfrac{1}{cos^2a}\)
=>\(\dfrac{1}{cos^2a}=1+9=10\)
=>\(cos^2a=\dfrac{1}{10}\)
\(B=\dfrac{3\cdot cosa-cosa}{27\cdot cos^3a+3\cdot cos^3a+2\cdot3\cdot cosa}\)
\(=\dfrac{2\cdot cosa}{30cos^3a+6cosa}=\dfrac{2}{30cos^2a+6}\)
\(=\dfrac{2}{3+6}=\dfrac{2}{9}\)
